3 \includegraphics[max width=\textwidth, alt={}, center]{f6887893-66c5-40df-ba8d-9439a5c268eb-3_351_314_255_918} A spring balance is modelled by a vertical light elastic spring \(A B\), of natural length 0.25 m and modulus of elasticity \(\lambda \mathrm { N }\). The bottom end \(B\) of the spring is fixed, and the top end \(A\) is attached to a small tray of mass 0.1 kg which is free to move vertically (see diagram). When in the equilibrium position, \(A B = 0.24 \mathrm {~m}\). Show that \(\lambda = 25\). The tray is pushed down by 0.02 m to the point \(C\) and released from rest. At time \(t\) seconds after release the displacement of the tray from the equilibrium position is \(x \mathrm {~m}\). Show that $$\ddot { x } = - 1000 x .$$ Find the time taken for the tray to move a distance of 0.03 m from \(C\).

## Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically at equilibrium: $0.1g = \lambda \cdot 0.01/0.25$ | M1 | |
| Evaluate $\lambda$: $\lambda = 25$ **A.G.** | A1 | |
| Use Newton's Law at general point: $0.1\,d^2x/dt^2 = 0.1g - \lambda(0.01 + x)/0.25$ | M1 | |
| Simplify: $d^2x/dt^2 = -1000x$ **A.G.** | A1 | |
| Use SHM formula for $x$: $x = 0.02\cos(t\sqrt{1000})$ [or $\sin$] | M1 A1 | |
| Find required time $t$: $t = (1/\sqrt{1000})\cos^{-1}(-0.01/0.02)$ | M1 A1 | |
| *or* $2\pi/4\sqrt{1000} + (1/\sqrt{1000})\sin^{-1}(1/2)$ | M1 A1 | |
| Evaluate: $t = (1/\sqrt{1000})\cdot 2\pi/3 = 0.0662$ [s] | A1 | **Total: 9** |

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\includegraphics[max width=\textwidth, alt={}, center]{f6887893-66c5-40df-ba8d-9439a5c268eb-3_351_314_255_918}

A spring balance is modelled by a vertical light elastic spring $A B$, of natural length 0.25 m and modulus of elasticity $\lambda \mathrm { N }$. The bottom end $B$ of the spring is fixed, and the top end $A$ is attached to a small tray of mass 0.1 kg which is free to move vertically (see diagram). When in the equilibrium position, $A B = 0.24 \mathrm {~m}$. Show that $\lambda = 25$.

The tray is pushed down by 0.02 m to the point $C$ and released from rest. At time $t$ seconds after release the displacement of the tray from the equilibrium position is $x \mathrm {~m}$. Show that

$$\ddot { x } = - 1000 x .$$

Find the time taken for the tray to move a distance of 0.03 m from $C$.

\hfill \mbox{\textit{CAIE FP2 2010 Q3 [9]}}